Example Question
a) Show that the equation x3 - 6x + 1 = 0 has a solution between 2 and 3.b) Solve this equation correct to 1 decimal place.
Solution
a) To show that x3 - 6x + 1 = 0 has a solution between 2 and 3 we need to substitute x = 2 and x = 3 into the equation.When x = 2 we get: (2)3 - 6(2) + 1 = 8 - 12 + 1 = -3.
When x = 3 we get: (3)3 - 6(3) + 1 = 27 - 6(3) + 1 = 10.
Now notice when x = 2 we get an answer less than 0 and when x = 3 we get an answer greater than 0. This means there must be a value of x between 2 and 3 which is equal to 0.
b) We now need to find this solution correct to 1 decimal place. To do this we need to draw a table to test different values of x. Since we know the answer is between 2 and 3 it makes sense to start with x = 2.5.
| x | x3 - 6x + 1 | Comment |
|---|---|---|
| 2 | -3 | Too small |
| 3 | 10 | Too big |
| 2.5 | 1.625 | Too big |
| 2.3 | -0.633 | Too small |
| 2.4 | 0.424 | Too big |
Now since 2.3 was too small and 2.4 was too big we know the solution is between these values. To find our answer to 1 decimal place we have to try one more value in the middle of these, when x = 2.35.
When x = 2.35 we get: (2.35)3 - 6(2.35) + 1 = -0.122125.
Finally since 2.35 is too small then we can say 2.3 must also be too small. Hence x = 2.4 to 1 decimal place.
Test Yourself!
The equation x3 + x = 15 has a solution between 2 and 3.Find this solution correct to one decimal place.
x =
The equation x3 - 2x + 6 = 0 has a solution between -3 and -2.
Find this solution correct to one decimal place.
x =